What is (\partial)/(\partial x)((x*y)/(x-y)) ?

The answer to (\partial)/(\partial x)((x*y)/(x-y)) is -(y^2)/((x-y)^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(\partial)/(\partial x)((x*y)/(x-y))

\frac{\partial}{\partial x} (\frac{x \cdot y}{x - y})

- \frac{y ^{2}}{( x - y ) ^{2}}

Solution steps

\frac{\partial}{\partial x} (\frac{x y}{x - y})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= y \frac{\partial}{\partial x} (\frac{x}{x - y})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= y \frac{\frac{\partial}{\partial x} ( x ) ( x - y ) - \frac{\partial}{\partial x} ( x - y ) x}{( x - y ) ^{2}}

\frac{\partial}{\partial x} (x) = 1

\frac{\partial}{\partial x} (x - y) = 1

= y \frac{1 \cdot ( x - y ) - 1 \cdot x}{( x - y ) ^{2}}

Simplify

y \frac{1 \cdot ( x - y ) - 1 \cdot x}{( x - y ) ^{2}} : - \frac{y ^{2}}{( x - y ) ^{2}}

= - \frac{y ^{2}}{( x - y ) ^{2}}

f (x) = - \frac{6 x}{x ^{2} + 1}, at x = 0

x \cdot cos (x)

\frac{\partial}{\partial x} (9 y e^{7 x})

\frac{x + 3}{x - 3}

\frac{\partial}{\partial y} (4 x + y)

What is (\partial)/(\partial x)((x*y)/(x-y)) ?
The answer to (\partial)/(\partial x)((x*y)/(x-y)) is -(y^2)/((x-y)^2)